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Question

(B) The force exerted by a fluid jet on a surface is given by the rate of change of momentum.Let $M$ be the mass of water hitting the surface in time

Vedclass · Accepted Answer

$mv^2$

Vedclass · Answer

(B) The force exerted by a fluid jet on a surface is given by the rate of change of momentum.Let $M$ be the mass of water hitting the surface in time $t$.The velocity of the water jet is $v$ and it comes to rest upon hitting the surface,so the change in velocity is $\Delta v = v - 0 = v$.The mass per unit length is given as $m = \frac{M}{L}$,where $L$ is the length of the water jet.In time $t$,the length of the water jet hitting the surface is $L = v \cdot t$.Therefore,the mass of water hitting the surface in time $t$ is $M = m \cdot L = m \cdot v \cdot t$.The force $F$ is the rate of change of momentum: $F = \frac{\Delta p}{\Delta t} = \frac{M \cdot v}{t}$.Substituting $M = m \cdot v \cdot t$ into the force equation:$F = \frac{(m \cdot v \cdot t) \cdot v}{t} = m \cdot v^2$.Thus,the force on the surface is $mv^2$.

An engine pumps water continuously through a hose. Water leaves the hose with velocity $v$ and $m$ is the mass per unit length of the water jet. If this jet hits a surface and comes to rest instantaneously,the force on the surface is

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